Tag: hyperbola

Questions Related to hyperbola

The second-degree curve and pair of asymptotes differ by a constant. Let the second-degree curve $S = 0$ represent the hyperbola then respective pair of asymptote is given by.$\displaystyle S+\lambda =0\left ( \lambda \in R \right )$ which represent a pair of straight lines so $\lambda$  can be determined. The equation of asymptotes is $\displaystyle A=s+\lambda =0$ if equation of conjugate hyperbola of the curve $S =0$ be represents by $S _{1}$, then $A$ is arithmetic mean of the curves $S _{1}$, & $ S $.

A hyperbola passing through origin has $\displaystyle 2x-y+3=0$ and $\displaystyle x-2y+2=0$ as its asymptotes, then equation of its transverse and conjugate axes are:

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $\displaystyle x-y+2=0$ and $\displaystyle 3x-3y+5=0$

  2. $\displaystyle x+y+2=0$ and $\displaystyle 3x-3y+5=0$

  3. $\displaystyle x-y+1=0$ and $\displaystyle 3x-3y+5=0$

  4. $\displaystyle2 x-2y+1=0$ and $\displaystyle 3x-3y+5=0$

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Correct Option: C
Explanation:

The transverse axis of hyperbola is the bisector of the angle between the asymptotes containing the origin and the conjugate axis is the other bisector.


 And equation of bisector of angle of the asymptotes are given by

$\displaystyle \frac{2x-y+3}{\sqrt{5}}=\pm \frac{x-2y+2}{\sqrt{5}}$

$\displaystyle \Rightarrow  2x-y+3 =\pm \left ( x-2y+2 \right )$

$\displaystyle \Rightarrow  2x-y+3 =x-2y+2$

and $\displaystyle  2x-y+3 =x-2y+2= -\left ( x-2y+2 \right )$

$\displaystyle  \Rightarrow x+y+1=0 \ and \ 3x-3y+5=0$

Hence, option 'C' is correct.

The asymptotes of a hyperbola have equations $y-1=\dfrac{3}{4}(x+3).$ If a focus of the hyperbola has coordinates $(7,1)$, the equation of the hyperbola is

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $\dfrac{(x+3)^2}{16}-\dfrac{(y-1)^2}{9} = 1$

  2. $\dfrac{(y-1)^2}{9}-\dfrac{(x+3)^2}{16} = 1$

  3. $\dfrac{(x+3)^2}{64}-\dfrac{(y-1)^2}{36} = 1$

  4. $\dfrac{(y-1)^2}{36}-\dfrac{(x+3)^2}{64} = 1$

  5. $\dfrac{(x+3)^2}{4}-\dfrac{(y-1)^2}{3} = 1$

Show answer
Correct Option: C
Explanation:

Equation of asymptotes are 

$y-1=\dfrac { 3 }{ 4 } (x+3  )$    ......(i)

$ y-1=-\dfrac { 3 }{ 4 } (x+3)$     .....(ii)

Centre of the hyperbola is point of intersection of asymptotes.

Therefore, by solving (i) and (ii), we get centre as $C(-3,1)$.

Slope of asymptotes $=\dfrac { b }{ a } $

$\Rightarrow \dfrac { b }{ a } =\pm \dfrac { 3 }{ 4 }$      ......(i)

Focus is $(7,1)$.

Focus for hyperbola of form $\dfrac { { (x-h) }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { (y-k) }^{ 2 } }{ { b }^{ 2 } } =1$ is $(h+ae,k)$

$\Rightarrow 7=-3+ae\\ \Rightarrow ae=10\\ \Rightarrow a\dfrac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  }{ a } =10\\ \Rightarrow \sqrt { { a }^{ 2 }+{ b }^{ 2 } } =10$

Substituting $b$ from (i), we get

$\Rightarrow \sqrt { { a }^{ 2 }+{ \left( \pm a \dfrac { 3 }{ 4 }  \right)  }^{ 2 } } =10\\ \Rightarrow \dfrac { 5a }{ 4 } =10\\ \Rightarrow a=8\\ \Rightarrow b=\pm \dfrac { 3 }{ 4 } a=\pm 6$

So, the equation of hyperbola is

$\dfrac { { (x+3) }^{ 2 } }{ { 8 }^{ 2 } } -\dfrac { { (y-1) }^{ 2 } }{ { 6 }^{ 2 } } =1$

$\dfrac { { (x+3) }^{ 2 } }{ 64 } -\dfrac { { (y-1) }^{ 2 } }{ 36 } =1$

So, option C is correct.

If $PN$ is the perpendicular from a point on a rectangular hyperbola to its asymptotes, the locus, then the midpoint of $PN$ is

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. circle

  2. parabola

  3. ellipse

  4. hyperbola

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Correct Option: D
Explanation:

Let $xy={ c }^{ 2 }$ be the rectangular hyperbola and let $P\left( { x
} _{ 1 },{ y } _{ 1 } \right) $ be apoint on it. Let $Q(h.k)$ be the
midpoint of $PN$. Then the coordinates of $Q$ are $\left( { { x } _{ 1
},{ y } _{ 1 } }/{ 2 } \right) $
$\therefore \quad { x } _{ 1 }={ h
}\quad \cfrac { { y } _{ 1 } }{ 2 } =k\Rightarrow { x } _{ 1 }={ h }\quad
,\quad { y } _{ 1 }=2k$
But $\left( { x } _{ 1 },{ y } _{ 1 } \right) $ lies on $xy={ c }^{ 2 }$
$\therefore \quad h(2k)={ c }^{ 2 }\Rightarrow hk=\cfrac { { c }^{ 2 } }{ 2 } $
Therefore, the locus of $(h,k)$ is $xy=\cfrac { { c }^{ 2 } }{ 2 } $, which is a hyperbola.
Hence, option 'D' is correct.

The asymptotes of the hyperbola $xy - 3x + 4y + 2 = 0$ are

mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

  1. $x= - 4$

  2. $x= 4$

  3. $y= - 3$

  4. $y= 3$

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Correct Option: B,D
Explanation:
Given : Hyperbola,
$xy-3x+4y+2=0$---------------1
For Asymptotes,
Let the Asymptote's Equation be $y=mx+c$
And then finding $\phi _{n}(m)$ by replacing $y\rightarrow m$ and $x\rightarrow 1$
As $n=2$,
$\phi _{2}(m)=m$
putting $\phi _{2}(m)=0$, we get $m=0$
By taking $m=0$, we will get only one asymptote parallel to X-axis, so let's find them with putting the co-efficients of higher terms to zero.
For Asymptote parallel to X-axis, we put co-efficient of highest degree of x to zero that is here 1, so co-efficient of x$=0$
$\Rightarrow (y-3)=0$------------2(from Equation 1)
For Asymptote parallel to Y-axis, we put co-efficient of highest degree of y to zero which is 1 here, co-efficient of y$=0    (from Equation 1)
$\Rightarrow x+4=0$------------3
The Equation 2 & 3 are asymptotes to Equation 1.
$x+4=0$ & $y-3=0$




Find the locus of the point of intersection of the lines $\sqrt 3 x-y-4\sqrt 3\lambda=0$ and $\sqrt 3 \lambda x +\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $3x^2-y^2=48$

  2. $y^2-3x^2=24$

  3. $4x^2-3y^2=16$

  4. None of these

Show answer
Correct Option: A
Explanation:

Let $(h,k)$ be the point of intersection of the given lines. Then,

$\sqrt 3 h-k-4\sqrt 3 \lambda=0$ and $\sqrt3 \lambda h +\lambda k-4\sqrt 3=0$
$\sqrt 3 h-k=4\sqrt 3\lambda$ and $\lambda(\sqrt 3h+k)=4\sqrt 3$
$(\sqrt 3 h-k)\lambda (\sqrt 3h +k)=(4\sqrt 3\lambda)(4\sqrt 3)$
$3h^2-k^2=48$
Hence, the locus of $(h,k)$ is $3x^2-y^2=48$.

If the equation of a hyperbola is $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{16}} = 1$, then 

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. traverse axis is along x-axis of length $6$

  2. traverse axis is along y-axis of length $8$

  3. conjugate axis is along y-axis of length $6$

  4. None of these

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Correct Option: A

The length of the transverse axis of the hyperbola $3x^2-4y^2=3$ is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\frac{{8\sqrt 2 }}{{\sqrt 3 }}$

  2. $\frac{{16\sqrt 2 }}{{\sqrt 3 }}$

  3. $\frac {3}{32}$

  4. $\frac {64}{3}$

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Correct Option: A

If the eccentricity and length of latus rectum of a hyperbola are $\frac {\sqrt 13}{3}$ and $\frac {10}{3}$ units respectively, then what is the length of the traverse axis?

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\frac {7}{2}$ units

  2. $12$ units

  3. $\frac {15}{2}$ units

  4. $\frac {15}{4}$ units

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Correct Option: A

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$ centre is

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $(4,4)$

  2. $(5,5)$

  3. $(4,5)$

  4. $(0,0)$

Show answer
Correct Option: D
Explanation:

Centre of the hyperbola of form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ has centre at orgin.

Here it is of this form. 
So, centre is at origin $(0,0)$.

For hyperbola  $\dfrac{x^2}{16}-\dfrac{y^2}{25}=1$ distances between two directrices are

mathematics and statistics hyperbola parametric equation of the hyperbola forms of equations of a hyperbola equations of hyperbola

  1. $\dfrac{16}{\sqrt{41}}$

  2. $\dfrac{25}{\sqrt{41}}$

  3. $\pm\dfrac{32}{\sqrt{141}}$

  4. $\dfrac{32}{\sqrt{41}}$

Show answer
Correct Option: D
Explanation:

Given equation of hyperbola $\dfrac {x^2}{16}-\dfrac {y^2}{25}=1$
Here $a^2=16, b^2=25$
Distance between directrix is  $\dfrac{2a^2}{\sqrt{a^2+b^2}} = \dfrac{2\times16}{\sqrt{16+25}} = \dfrac{32}{\sqrt{41}}$